Question: Divide the following complex numbers. $ \dfrac{-8-9i}{-2-i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2+i}$ $ \dfrac{-8-9i}{-2-i} = \dfrac{-8-9i}{-2-i} \cdot \dfrac{{-2+i}}{{-2+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8-9i) \cdot (-2+i)} {(-2-i) \cdot (-2+i)} = \dfrac{(-8-9i) \cdot (-2+i)} {(-2)^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8-9i) \cdot (-2+i)} {(-2)^2 - (-1i)^2} = $ $ \dfrac{(-8-9i) \cdot (-2+i)} {4 + 1} = $ $ \dfrac{(-8-9i) \cdot (-2+i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-9i}) \cdot ({-2+i})} {5} = $ $ \dfrac{{-8} \cdot {(-2)} + {-9} \cdot {(-2) i} + {-8} \cdot {1 i} + {-9} \cdot {1 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{16 + 18i - 8i - 9 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{16 + 18i - 8i + 9} {5} = \dfrac{25 + 10i} {5} = 5+2i $